
What is ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$?
Answer
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Hint: Here we will try to find the value of angle for which its cosine value is $\dfrac{1}{2}$ in the range of angle $\left[ 0,\pi \right]$. We will assume this angle as $\theta $ and write the above expression as ${{\cos }^{-1}}\left( \cos \theta \right)$. Now, we will use the formula ${{\cos }^{-1}}\left( \cos \theta \right)=\theta $ to get the value of $\theta $ as the answer.
Complete step by step solution:
Here we have been provided with the inverse trigonometric expression ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ and we are asked to find its value. As it is an inverse trigonometric function so the value which we will find will be called as the principal value and this value is going to be an angle.
We know that the range of inverse cosine function $\left( {{\cos }^{-1}}x \right)$ is $\left[ 0,\pi \right]$. So the expression ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ means we have to select a such value of the angle that must lie in the range $\left[ 0,\pi \right]$ and the cosine value of this angle is \[\dfrac{1}{2}\].
Now, we know that the value of cosine function is \[\dfrac{1}{2}\] when the angle Is $\dfrac{\pi }{3}$ which clearly lies in the range $\left[ 0,\pi \right]$. Therefore, the expression ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ can be written as:
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)={{\cos }^{-1}}\left( \cos \dfrac{\pi }{3} \right)$
We know that ${{\cos }^{-1}}\left( \cos x \right)=x$ when ‘x’ lies in the range of angle $\left[ 0,\pi \right]$ therefore we can simplify the above expression as:
$\begin{align}
& \Rightarrow {{\cos }^{-1}}\left( \cos \dfrac{\pi }{3} \right)=\dfrac{\pi }{3} \\
& \therefore {{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3} \\
\end{align}$
Hence, the principal value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ is $\dfrac{\pi }{3}$.
Note: You may note that there is only one principal value of an inverse trigonometric function. There are infinite angles for which we will get the value of cosine function equal to $\dfrac{1}{2}$ but we have to remember the range in which cosine inverse function is defined and we need to choose the angle from that range only. Also remember that the domain of inverse cosine function is $x\in \left[ -1,1 \right]$, which is actually the range of the cosine function.
Complete step by step solution:
Here we have been provided with the inverse trigonometric expression ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ and we are asked to find its value. As it is an inverse trigonometric function so the value which we will find will be called as the principal value and this value is going to be an angle.
We know that the range of inverse cosine function $\left( {{\cos }^{-1}}x \right)$ is $\left[ 0,\pi \right]$. So the expression ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ means we have to select a such value of the angle that must lie in the range $\left[ 0,\pi \right]$ and the cosine value of this angle is \[\dfrac{1}{2}\].
Now, we know that the value of cosine function is \[\dfrac{1}{2}\] when the angle Is $\dfrac{\pi }{3}$ which clearly lies in the range $\left[ 0,\pi \right]$. Therefore, the expression ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ can be written as:
$\Rightarrow {{\cos }^{-1}}\left( \dfrac{1}{2} \right)={{\cos }^{-1}}\left( \cos \dfrac{\pi }{3} \right)$
We know that ${{\cos }^{-1}}\left( \cos x \right)=x$ when ‘x’ lies in the range of angle $\left[ 0,\pi \right]$ therefore we can simplify the above expression as:
$\begin{align}
& \Rightarrow {{\cos }^{-1}}\left( \cos \dfrac{\pi }{3} \right)=\dfrac{\pi }{3} \\
& \therefore {{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3} \\
\end{align}$
Hence, the principal value of ${{\cos }^{-1}}\left( \dfrac{1}{2} \right)$ is $\dfrac{\pi }{3}$.
Note: You may note that there is only one principal value of an inverse trigonometric function. There are infinite angles for which we will get the value of cosine function equal to $\dfrac{1}{2}$ but we have to remember the range in which cosine inverse function is defined and we need to choose the angle from that range only. Also remember that the domain of inverse cosine function is $x\in \left[ -1,1 \right]$, which is actually the range of the cosine function.
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